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volume 7, issue 5, article 179, 2006.

Received 15 May, 2006;

accepted 13 July, 2006.

Communicated by:J. Sándor

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Journal of Inequalities in Pure and Applied Mathematics

SOME INEQUALITIES INVOLVING THE GAMMA FUNCTION

LAZHAR BOUGOFFA

Al-imam Muhammad Ibn Saud Islamic University Faculty of Computer Science

Department of Mathematics P.O. Box 84880, Riyadh 11681 Saudi Arabia

EMail:bougoffa@hotmail.com

c

2000Victoria University ISSN (electronic): 1443-5756 142-06

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Some Inequalities Involving the Gamma Function

Lazhar Bougoffa

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J. Ineq. Pure and Appl. Math. 7(5) Art. 179, 2006

http://jipam.vu.edu.au

Abstract

In this short paper, as a complement of the double inequality on the Euler gamma function, obtained by József Sándor in the paper [A note on certain inequalities for the gamma function,J. Ineq. Pure Appl. Math.,6(3) (2005), Art.

61], several inequalities involving the Euler gamma function are established by using the same method of J. Sándor that is used in [2].

2000 Mathematics Subject Classification:33B15.

Key words: Euler gamma function, Digamma function.

1. Introduction and Lemma

In [1], C. Alsina and M.S. Tomas studied a very interesting inequality involving the gamma function and proved the following double inequality

(1.1) 1

n! ≤ Γ(1 +x)ⁿ

Γ(1 +nx) ≤1, x∈[0,1], n∈ ℵ,

by using a geometrical method [1]. In view of the interest in this type of inequalities, J. Sándor [2] extended this result to a more general case, and obtained the following inequality

(1.2) 1

Γ(1 +a) ≤ Γ(1 +x)^a

Γ(1 +ax) ≤1, x∈[0,1], a≥1.

The method used in [2] to obtain these results is based on the following lemma.

Lemma 1.1. For allx >0, and alla≥1one has

(1.3) ψ(1 +ax)≥ψ(1 +x),

where ψ(x) = ^Γ_Γ(x)⁰^(x), x > 0 is the digamma function and has the following series representation

(1.4) ψ(x) =−γ+ (x−1)

∞

X

k=0

1

(k+ 1)(x+k). Proof. See [2].

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In [3], A.McD. Mercer continued to create new inequalities on this subject and other special functions and obtained the following inequalities

(1.5) Γ(1 +x)^a

Γ(1 +ax) < Γ(1 +y)^a

Γ(1 +ay), 0< a <1,

(1.6) Γ(1 +x)^a

Γ(1 +ax) > Γ(1 +y)^a

Γ(1 +ay), a <0ora >1, wherey > x >0,1 +ax >0, and1 +bx >0.

This paper is a continuation of the above papers. As in [2], our goal is to prove several inequalities involving the gamma function, using the same method of J. Sándor that is used in [2]. Here, the essential lemma is the following Lemma 1.2. For allx >0, and alla≥bwe have

(1.7) ψ(1 +ax)≥ψ(1 +bx),

in which1 +ax >0and1 +bx >0.

Proof. By the above series representation ofψ, observe that:

ψ(1 +ax)−ψ(1 +bx) =

∞

X

k=0

ax

(k+ 1)(ax+k+ 1) − bx

(k+ 1)(bx+k+ 1)

,

ψ(1 +ax)−ψ(1 +bx) = (a−b)x

∞

X

k=0

1

(ax+k+ 1)(bx+k+ 1) ≥0, bya ≥b,1 +ax >0,1 +bx >0,x >0andk >0. Thus the inequality (1.7) is proved. The equality in (1.7) holds only ifa=b.

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2. Main Results

Now we are in a position to give the following theorem.

Theorem 2.1. Letf be a function defined by

f(x) = Γ(1 +bx)^a

Γ(1 +ax)^b, ∀x≥0,

in which 1 +ax > 0and 1 +bx > 0, then for all a ≥ b > 0 or0 > a ≥ b (a >0andb <0),f is decreasing (increasing) respectively on[0,∞).

Proof. Letgbe a function defined by

g(x) = logf(x) =alog Γ(1 +bx)−blog Γ(1 +ax), then

g⁰(x) =ab[ψ(1 +bx)−ψ(1 +ax)].

By Lemma 1.2, we get g⁰(x) ≤ 0if a ≥ b > 0 or 0 > a ≥ b (g⁰(x) ≥ 0if 0 > a ≥ b), i.e.,g is decreasing on[0,∞)(increasing on[0,∞)) respectively.

Hencef is decreasing on[0,∞)ifa≥b >0or0> a≥b(increasing ifa >0 andb <0) respectively.

The proof is complete.

Corollary 2.2. For allx∈[0,1], and alla ≥b >0or0> a≥b, we have

(2.1) Γ(1 +b)^a

Γ(1 +a)^b ≤ Γ(1 +bx)^a Γ(1 +ax)^b ≤1.

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Proof. To prove (2.1), applying Theorem2.1, and taking account of Γ(1) = 1 we getf(1)≤f(x)≤f(0)for allx∈[0,1], and we omit (2.1).

Corollary 2.3. For allx∈[0,1], and alla >0andb <0, we have

(2.2) 1≤ Γ(1 +bx)^a

Γ(1 +ax)^b ≤ Γ(1 +b)^a Γ(1 +a)^b.

Proof. Applying Theorem 2.1, we get f(0) ≤ f(x) ≤ f(1) for allx ∈ [0,1], and we omit (2.2).

Now we consider the simplest cases of Corollary 2.2 to obtain the known results of C. Alsina and M.S. Tomas [1] and J. Sándor [2].

Remark 1. Taking a = nand b = 1(a ≥ 1andb = 1), in Corollary2.2, we obtain (1.1) ((1.2)) respectively.

Also we conclude different generalizations of (1.5)–(1.6) which are obtained by A.McD. Mercer [3].

Corollary 2.4. For allx∈[0,1], and alla ≥b >0or0> a≥b, we have

(2.3) Γ(1 +bx)^a

Γ(1 +ax)^b < Γ(1 +by)^a Γ(1 +ay)^b, where0< y < x≤1.

Corollary 2.5. For allx∈[0,1], and alla >0andb <0, we have

(2.4) Γ(1 +by)^a

Γ(1 +ay)^b < Γ(1 +bx)^a Γ(1 +ax)^b, where0< y < x≤1.

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References

[1] C. ALSINA AND M.S. TOMAS, A geometrical proof of a new inequal- ity for the gamma function, J. Ineq. Pure Appl. Math., 6(2) (2005), Art.

48. [ONLINE:http://jipam.vu.edu.au/article.php?side=

517].

[2] J. SÁNDOR, A note on certain inequalities for the gamma function, J. Ineq.

Pure Appl. Math., 6(3) (2005), Art. 61. [ONLINE:http://jipam.vu.

edu.au/article.php?side=534].

[3] A.McD. MERCER, Some new inequalities for the gamma, beta and zeta functions, J. Ineq. Pure Appl. Math., 7(1) (2006), Art. 29. [ONLINE:

http://jipam.vu.edu.au/article.php?side=636].